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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2024 · 4 April · Morning Shift · Physics · Atomic Physics

A hydrogen atom changes its state from n=3 to n=2 . Due to recoil,...

A hydrogen atom changes its state from n=3n=3 to n=2n=2. Due to recoil, the percentage change in the wave length of emitted light is approximately 1×10−n1 \times 10^{-n}. The value of nn is_____. [Given Rhc=13.6eV,hc=1242eVnm,h=6.6×10−34 J s\mathrm{Rhc}=13.6 \mathrm{eV}, \mathrm{hc}=1242 \mathrm{eV} \mathrm{nm}, \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s} mass of the hydrogenatom =1.6×10−27 kg=1.6 \times 10^{-27} \mathrm{~kg} ]

Correct answer

0

Step-by-step solution

ΔE=13.6(122−132)=1.9eVΔE=hcλλ=hcΔEPi=Pt0=−mv+hλ′′⇒v=hmλ′ΔE=12mv2+hcλ′′=12m(hmλ′)2+hcλ′\begin{aligned} & \Delta \mathrm{E}=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=1.9 \mathrm{eV} \\ & \Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda} \\ & \lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}} \\ & \mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{t}} \\ & 0=-\mathrm{mv}+\frac{\mathrm{h}}{\lambda^{\prime \prime}} \\ & \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}} \\ & \Delta \mathrm{E}=\frac{1}{2} \mathrm{mv^{2 }}+\frac{\mathrm{hc}}{\lambda^{\prime \prime}} \\ & =\frac{1}{2} \mathrm{ m}\left(\frac{\mathrm{h}}{\mathrm{m} \lambda^{\prime}}\right)^2+\frac{\mathrm{hc}}{\lambda^{\prime}} \end{aligned}Now ΔE=h22mλ′2+hcλ′λ′2ΔE−hcλ′−h22m=0λ′=hc±h2c2+4ΔEh22m2ΔEλ′=hc±hc1+2ΔEmc2ΔE\begin{aligned} & \Delta \mathrm{E}=\frac{h^2}{2 \mathrm{ m} \lambda^{\prime^2}}+\frac{\mathrm{hc}}{\lambda^{\prime}} \\ & \lambda^{\prime 2} \Delta \mathrm{E}-\mathrm{hc} \lambda^{\prime}-\frac{\mathrm{h}^2}{2 \mathrm{ m}}=0 \\ & \lambda^{\prime}=\frac{\mathrm{hc} \pm \sqrt{\mathrm{h}^2 \mathrm{c}^2+\frac{4 \Delta \mathrm{Eh}^2}{2 \mathrm{ m}}}}{2 \Delta \mathrm{E}} \\ & \lambda^{\prime}=\frac{\mathrm{hc} \pm \mathrm{hc} \sqrt{1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}}}}{2 \Delta \mathrm{E}} \end{aligned}λ′λ=1+(1+2ΔEmc2)122=1+1+ΔEmc22λ′λ=1+ΔE2mc2λ′−λλ=ΔE2mc2=1.9×1.6×10−192×1.67×10−27×9×1015=10−9∴% change ≈10−7\begin{aligned} & \frac{\lambda^{\prime}}{\lambda}=\frac{1+\left(1+\frac{2 \Delta \mathrm{E}}{\mathrm{mc}^2}\right)^{\frac{1}{2}}}{2}=\frac{1+1+\frac{\Delta \mathrm{E}}{\mathrm{mc}^2}}{2} \\ & \frac{\lambda^{\prime}}{\lambda}=1+\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^2} \\ & \frac{\lambda^{\prime}-\lambda}{\lambda}=\frac{\Delta \mathrm{E}}{2 \mathrm{mc}^2}=\frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times 9 \times 10^{15}}=10^{-9} \\ & \therefore \% \text{ change } \approx 10^{-7} \end{aligned}$Correct answer 7
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